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Take the Dirichlet Problem as example:

The difinition of Dirichlet Problem from wiki

Given a function $f$ that has values everywhere on the boundary of a region in $\mathbb{R}^n$, is there a unique continuous function $u$ twice continuously differentiable in the interior and continuous on the boundary, such that $u$ is harmonic in the interior and $u=f$ on the boundary?

Does 'harmonic in the interior' implicitly and necessarily means that 'harmonic on an open superset of the closed region'?

Should I interpret a PDE problem this way? Namely, a PAE problem is actually defined on the open super set of the problem region?

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I do not understand: the solution is harmonic inside. Why should it be harmonic also outside? –  Siminore Sep 20 '12 at 16:40
    
For the Dirichlet Problem on Unit Disc, let f=0 on the Unit Circle. Let $P_r(\theta)=(1-r^2)/(1-2r\cos{\theta}+r^2)$ be the Poisson kernel, then define $u$ by $u(r,\theta ) = \partial Pr/\partial \theta ,\ 0 \le r < 1;\ u=0,\ r=1$, then $u$ is harmonic in the disc ($\Delta u=0$) and continuous on the boundary($\lim_{r\rightarrow 1}u(r,\theta)=0$), but $u$ does not solve the Dirichlet Problem($u$ is not identically zero). –  xoofee Sep 21 '12 at 2:34
    
(continuing...) I think the reason is $u$ is not harmonic on any open superset of the closed disc, that is to say, $u$ can't be extended to any harmonic function on any open super set of the closed disc. –  xoofee Sep 21 '12 at 2:35
    
There's something strange about this discussion. Why do you need $u=0$? –  Siminore Sep 21 '12 at 7:39
    
$f$ does not have to be continuous on the boundary. Usually one studies $f \in L^p(\partial \Omega)$. Sometimes you can even talk about even less regular $f$ than that, eg, $f$ in negative sobolev spaces. The wiki statement is more restrictive than what people usually mean when they talk about a Dirichlet problem. –  Nick Alger Sep 21 '12 at 15:07

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The answer is no: there is no implicit assumption that $u$ is harmonic on a superset of the closure. In fact, the boundary data $f$ may well be some kind of continuous but nowhere differentiable function like Weierstrass's. That would obviously preclude any possibility of harmonic extension of $u$ to a larger domain.

Added I think I see the source of confusion. You are right that the function $u=\partial P/\partial \theta $ has radial limit zero at every point of the boundary. But this does not mean that $u$ is continuous on the closed disk. Indeed, it is not even bounded on the disk. The Dirichlet problem (with continuous boundary data) requires continuous extension to the boundary, not just the existence of radial limits.

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See my above comment, $u$ does not solve the unit disc dirichlet problem, why? (It comes from Stein's Fourier Analysis, page.64, Chapter 2, Exercise 18.) –  xoofee Sep 21 '12 at 6:59
    
Prof Jerry Shurman's lecture: DIRICHLET'S PROBLEM ON THE DISK third paragraph on the first page: More precisely, u is defined on an open superset $\Omega$ of $\overline{D}$ what do you say to that? –  xoofee Sep 21 '12 at 7:07
    
@xoofee Poisson kernel is not continuous in the closed disk. Neither are its derivatives. This is why they do not solve the Dirichlet problem with boundary value $\equiv0$. –  user31373 Sep 21 '12 at 14:23
    
@xoofee: Look at the same lecture, page 3, paragraph 2: what do you say to that? –  timur Sep 21 '12 at 15:01
    
I repeated here: define $u$ by $u(r,\theta ) = \partial Pr/\partial \theta ,\ 0 \le r < 1;\ u=0,\ r=1$, then $u$ is harmonic in the disc ($\Delta u=0$) and continuous on the boundary($\lim_{r\rightarrow 1}u(r,\theta)=0$), but $u$ does not solve the Dirichlet Problem, what wrong? –  xoofee Sep 21 '12 at 15:01

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